Let f(x) be a polynomial such that f(f(x))−x^2 =xf(x). Find f(-100)
If the degree of the polynomial f(x) is d, then the degree of f(f(x)) is d^2. If d > 1, then f(f(x)) - x^2 has degree d^2, while x f(x) has degree d+1. But d^2 = d+1 has no integer solutions. Therefore d must be equal to 1, this means that:
f(x) = a x + b
f(f(x)) - x^2 =
a (a x + b) + b -x^2 =
a^2 x + a b + b -x^2
This has to be equal to
x f(x) = a x^2 + b x
for all x, therefore:
a = -1
b = a^2 = 1
a b + b must be equal to zero, and this indeed the case.
So, f(x) = -x + 1.
thanks.......
To find the value of f(-100), we need to solve the given equation and obtain the polynomial f(x).
Let's work through the equation step by step:
f(f(x))-x^2 = x*f(x)
First, let's substitute x = -100 into the equation:
f(f(-100)) - (-100)^2 = (-100) * f(-100)
Simplifying further:
f(f(-100)) - 10000 = -100 * f(-100)
Now, let's denote f(-100) as a new variable, y:
f(y) - 10000 = -100y
Rearranging the equation to isolate f(y):
f(y) + 100y - 10000 = 0
We have obtained a polynomial equation in terms of y. To find its roots, we can use factoring, the quadratic formula, or any other method for solving polynomial equations.
Solving the equation, we find that the roots are y = 100 and y = -100.
Thus, f(-100) can be either 100 or -100. Without further information or constraints, we cannot determine the exact value of f(-100).